Problem: The scalar field $f(x, y) = x\sin(y) + (y - 1)^2$ has a critical point at $(2, 0)$. How does the second partial derivative test classify this critical point? Choose 1 answer: Choose 1 answer: (Choice A) A Local maximum (Choice B) B Local minimum (Choice C) C Saddle point (Choice D) D The test is inconclusive
Explanation: The second partial derivative test uses the quantity below, evaluated at the critical point we wish to classify. $H = f_{xx}f_{yy} - f_{xy}f_{yx}$ $H < 0$ implies a saddle point. $H > 0$ and $f_{xx} > 0$ implies a local minimum. $H > 0$ and $f_{xx} < 0$ implies a local minimum. $H = 0$ means the test is inconclusive. Let's calculate $H$. First we need all the regular partial derivatives. $\begin{aligned} f_x &= \sin(y) \\ \\ f_y &= x\cos(y) + 2(y - 1) \end{aligned}$ Now we can find all the second order partial derivatives. $\begin{aligned} f_{xx} &= 0 \\ \\ f_{yx} &= \cos(y) = 1 \\ \\ f_{xy} &= \cos(y) = 1 \\ \\ f_{yy} &= -x\sin(y) + 2 = 2 \end{aligned}$ Therefore, $H = (0)(2) - (1)(1) = -1$. Because $H$ is negative, we know that the critical point is a saddle point.